101120300113 has 4 divisors (see below), whose sum is σ = 115566057280. Its totient is φ = 86674542948.

The previous prime is 101120300107. The next prime is 101120300117. The reversal of 101120300113 is 311003021101.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also an emirpimes, since its reverse is a distinct semiprime: 311003021101 = 1151 ⋅270202451.

It is not a de Polignac number, because 101120300113 - 2¹⁷ = 101120169041 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 101120300093 and 101120300102.

It is not an unprimeable number, because it can be changed into a prime (101120300117) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 7222878573 + ... + 7222878586.

It is an arithmetic number, because the mean of its divisors is an integer number (28891514320).

Almost surely, 2^101120300113 is an apocalyptic number.

It is an amenable number.

101120300113 is a deficient number, since it is larger than the sum of its proper divisors (14445757167).

101120300113 is an equidigital number, since it uses as much as digits as its factorization.

101120300113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 14445757166.

The product of its (nonzero) digits is 18, while the sum is 13.

Adding to 101120300113 its reverse (311003021101), we get a palindrome (412123321214).

The spelling of 101120300113 in words is "one hundred one billion, one hundred twenty million, three hundred thousand, one hundred thirteen".